Category: Class 11 Biology

NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-20-locomotion-movement/

NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement

Multiple Choice Questions

Q1. Match the following and mark the correct option.

Column I		Column II
A.	Fast muscle fibres	(i)	Myoglobin
B.	Slow muscle fibres	(ii)	Lactic acid
C.	Actin filament	(iii)	Contractile unit
D.	Sarcomere	(iv)	I-band

Options:
(a) A—(i), B—(ii), C—(iv), D—(iii)
(b) A—(ii), B—(i), C—(iii), D—(iv)
(c) A—(ii), B—(i), C—(iv), D—(iii)
(d) A—(iii), B—(ii), C—(iv), D—(i)

Ans. (c)

A.	Fast muscle fibres	(ii)	Lactic acid
B.	Slow muscle fibres	(i)	Myoglobin
C.	Actin filament	(iv)	I-band
D.	Sarcomere	(iii)	Contractile unit

Q2. Ribs are attached to
(a) Scapula (b) Sternum (c) Clavicle (d) Ilium
Ans: (b) Ribs are attached to sternum.

Q3. What is the type of movable joint present between the atlas and axis?
(a) Pivot (b) Saddle (c) Hinge (d) Gliding
Ans: (a) Pivot joint: Between atlas and axis called atlanto-axial joint.

Q4. ATPase of the muscle is located in
(a) Actinin (b) Troponin (c) Myosin (d) Actin
Ans: (c) ATPase of the muscle is located in head of myosin.

Q5. Intervertebral disc is found in the vertebral column of
(a) Birds (b) Reptiles (c) Mammals (d) Amphibians
Ans: (c) Intervertebral disc is found in the vertebral column of mammals.

Q6. Which one of the following is showing the correct sequential order of vertebrae in the vertebral column of human beings? ‘
(a) Cervical — lumbar — thoracic — sacral — coccygeal
(b) Cervical — thoracic — sacral — lumbar — coccygeal
(c) Cervical — sacral — thoracic — lumbar — coccygeal
(d) Cervical — thoracic — lumbar — sacral — coccygeal
Ans: (d) Cervical—thoracic—lumbar—sacral—coccygeal is the correct sequential order of vertebrae in the vertebral column of human beings.

Q7. Which one of the following options is incorrect?
(a) Hinge joint—between humerus and pectoral girdle
(b) Pivot joint—between atlas, axis and occipital condyle
(c) Gliding joint—between the carpals
(d) Saddle joint—between carpel and metacarpals of thumb
Ans: (a) Hinge joint—Knee joint and elbow joint

Q8. Knee joint and elbow joints are examples of
(a) Saddle joint (b) Ball and socket joint
(c) Pivot joint (d) Hinge joint
Ans: (d) Knee joint and elbow joints are examples of hinge joint.

Q9. Macrophages and leucocytes exhibit
(a) Ciliary movement
(b) Flagellar movement
(c) Amoeboid movement
(d) Gliding movement
Ans: (c) Amoeboid movements: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

Q10. Which one of the following is not a disorder of bone?
(a) Arthritis
(b) Osteoporosis
(c) Rickets
(d) Atherosclerosis
Ans: (d) Atherosclerosis is a disorder of circulatory system.

Q11. Which one of the following statement is incorrect?
(a) Heart muscles are striated and involuntary
(b) The muscles of hands and legs are striated and voluntary
(c) The muscles located in the inner walls of alimentary canal are striated and involuntary
(d) Muscles located in the reproductive tracts are unstriated and involuntary
Ans:(c) The muscles located in the inner walls of alimentary canal are non- striated and involuntary.

Q12. Which one of the following statements is-true?
(a) Head of humerus bone articulates with acetabulum of pectoral girdle
(b) Head of humerus bone articulates with glenoid cavity of pectoral girdle
(c) Head of humerus bone articulates with a cavity called acetabulum of pelvic girdle
(d) Head of humerus bone articulates with a glenoid cavity of pelvic girdle
Ans: (b) Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

Q13. Muscles with characteristic striations and involuntary are
(a) Muscles in the wall of alimentary canal
(b) Muscles of the heart
(c) Muscles assisting locomotion
(d) Muscles of the eyelids
Ans: (b) Muscles with characteristic striations and involuntary are muscles of the heart (Cardiac muscles).

Q14. Match the followings and mark the correct option.

Column I		.                  Column II
A.	Sternum	(i)	Synovial fluid
B.	Glenoid cavity	(ii)	Vertebrae
C.	Freely movable joint	(iii)	Pectoral girdle
D.	Cartilaginous joint	(iv)	Flat bones

Options:

(a) A—(ii), B—(i), C—{iii), D—(iv)
(b) A—(iv), B—(iii), C—(i), D—(ii)
(c) A—(ii), B—(i), C—(iv), D—(iii)
(d) A—(iii), B—(i), C—(ii), D—(iv)

Ans. (b)

A.	Sternum	(iv)	Flat bones
B.	Glenoid cavity	(iii)	Synovial fluid
C.	Freely movable joint	(i)	Pectoral girdle
D.	Cartilaginous joint	(ii)	Yertebrae

 

Very Short Answer Type Questions
Q1. Name the cells/tissues in human body which
a. exhibit amoeboid movement
b. exhibit ciliary movement
Ans: a. Macrophages and leucocytes
b. Trachea, fallopian tube and bronchiole

Q2. Locomotion requires a perfect‘coordinated activity of muscular _____________, _______ , systems.
Ans: Skeletal and Neural

Q3. Sarcolemma, sarcoplasm and sarcoplasmic reticulum refer to a particular type of cell in our body. Which is this cell and to what parts of that cell do these names refer to?
Ans: Each muscle fibre or muscle cell is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm. A muscle fibre is a syncytium as the sarcoplasm (cytoplasm) contains many nuclei. The endoplasmic reticulum, i. e., sarcoplasmic reticulum of the muscle fibres is the store house of calcium ions.

Q4 .Label the different components of actin filament in the diagram given

Q5. The. three tiny bones present in middle ear are called ear ossicles. Write them in correct sequence beginning from eardrum.
Ans: Malleus, incus and stapes.

Q6. What is the difference between the matrix of bones and cartilage?
Ans: Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength.
The inter-cellular material of cartilage is solid and pliable which resists compression. Cell of cartilage are called chondrocytes which are enclosed in a small cavities (lacunae) within the matrix secreted by them.

Q7. Which tissue is afflicted by Myasthenia gravis? What is the underlying cause?
Ans: Myasthenia gravis: Auto-immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.

Q8. How do our bone joints function without grinding noise and pain?
Ans: Our bone joints function without grinding noise and pain due to the presence of synovial fluid between bones.

Q9. Give the location of a ball and socket joint in a human body.
Ans: Ball and socket joint: Between humerus and pectoral girdle (shoulder joint). Between femur and acetabulum of pelvic girdle (hip joint). Total 4 ball and socket joints present in human body -2 shoulder joint and 2 hip joint.

Q10. Our forearm is made of three different bones. Comment.
Ans: The bones of the forearm are humerus, radius and ulna.

Q1. With respect to rib cage, explain the following:
a. Bicephalic ribs
b. True ribs
c. Floating ribs
Ans: a. Bicephalic ribs: Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic.
b. True ribs: First seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.
c. Floating ribs: Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are therefore, called floating ribs.

Q2. In old age, people often suffer from stiff and inflamed joints. What is this condition called? What are the possible reasons for these symptoms?
Ans: In old age, people suffer from stiff and inflamed joints, it is due to rheumatoid arthritis (autoimmune disorder)
Causes: (i) Inflammation of synovial membrane
(ii) Genetic factors (50% cases) .
(iii) Smoking
(iv) Vitamin-D deficiency

Q3. Exchange of calcium between bone and extracellular fluid takes place under the influence of certain hormones.

1. What will happen if more of Ca ⁺ is in extracellular fluid?
2. What will happen if very less amount of Ca⁺⁺ is in the extracellular fluid?

Ans: a. If more of Ca⁺⁺ is in extracellular fluid then it will be accumulated on the bones under the influence of thyrocalcitonin (TCT).
b. If very less amount of Ca⁺⁺ is in the extracellular fluid then parathyroid hormone (PTFI) acts on bones and stimulates the process of bone resorption (dissolution/demineralisation). PTH also stimulates reabsorption of Ca²⁺ by the renal tubules and increases Ca²⁺ absorption from the digested food.

Q4. Name at least two hormones which result in fluctuation of Ca⁺⁺

Ans: Thyrocalcitonin (TCT) and Parathyroid Hormone (PTH).

Q5. Rahul exercises regularly by visiting a gymnasium. Of late he is gaining weight. What could be the reason? Choose the correct answer and elaborate.
a. Rahul has gained weight due to accumulation of fats in body.
b. Rahul has gained weight due to increased muscle and less of fat.
c. Rahul has gained weight because his muscle shape has improved.
d. Rahul has gained weight bdcause he is accumulating water in the body.
Ans: Rahul has gained weight due to increased muscle and less of fat.

Q6. Radha was running on a treadmill at a great speed for 15 minutes continuously. She stopped the treadmill and abruptly came out. For the next few minutes, she was breathing heavily/fast. Answer the following questions:
a. What happened to her muscles when she did strenuously exercise?
b. How did her breathing rate change?
Ans: a. Repeated activation of the muscles can lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen in them, causing fatigue.
b. During strenuous exercise demand of oxygen also increases so breathing rate has been changed.

Q7. Write a few lines about Gout.
Ans: When metabolic waste-uric acid crystals are accumulated in bones, then it results into inflammation of bone and joints thereby causing pain. This disorder of skeletal system is called gout.

Q8. What is the source of energy for muscle contraction?
Ans: ATP (Adenosine Triphosphate)

Q9. What are the points for articulation of pelvic and pectoral girdles?
Ans: The components of pelvic girdle are ilium, ischium and pubis. It articulates with, femur through acetabulum. The components of pectoral girdle are scapula and clavicle. It is the glenoid cavity of pectoral girdle in which head . of humerus articulates.

Long Answer Type Questions

Q1. Calcium ion concentration in blood affects muscle contraction. Does it lead to tetany in certain cases? How will you correlate fluctuation in blood calcium with tetany?
Ans: Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching this junction releases a neurotransmitter (acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca⁺⁺ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross-bridge. This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. The process continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments.

Tetany: Rapid spasms (wild contractions) in muscle due to low Ca in body fluid.

Q2. An elderly woman slipped in th£ bathroom and had severe pain in her lower back. After X-ray examination doctors told her it is due to a slipped disc. What does that mean? How does it affect our health?
Ans: Displacement of intervertebral disc from’ their normal position is called slipped disc.
Effects:
i. Neck or lower back pain
ii. Muscular weakness
iii. Paralysis
iv. Sciatica

Q3. Explain sliding filament theory of muscle contraction with neat sketches.
Ans. Mechanism of muscle contraction: Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments. Muscle contraction is initiated by a signal sent by the Central Nervous System (CNS) via a motor neuron. A motor neuron alongwith the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca⁺⁺ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin.Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross-bridge. This pulls the attached

actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. It is clear from the above steps, that during shortening of the muscle, i.e., contraction, the ‘I’ bands get reduced, whereas the ‘A’ bands retain the length. The myosin, releasing the ADP and P, goes back to its relaxed state. A new ATP binds and the cross-bridge is broken. The ATP is again hydrolysed by the myosin head and the cycle of cross-bridge formation and breakage is repeated causing further sliding. The process continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic cistemae resulting in the masking of actin filaments. This causes the return of ‘Z’ lines back to their original position, i.e., relaxation.

Q5. Discuss the role of Ca²⁺ ions in muscle contraction. Draw neat sketches to illustrate your answer.
Ans: Muscle contraction is initiated by a neural signal, which after reaching neuromuscular junction or motor end plate releases a neurotransmitter, as a result an action potential in the sarcolemma is generated. Action potential spreads through muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca²⁺ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby removes the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active site on actin to form a cross-bridge. This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing shortening of the sarcomere, i.e., contraction.

A new ATP binds to myosin head and the cross-bridge is broken. The ATP is again hydrolysed by the myosin head and the cycle of cross-bridge formation and breakage is repeated causing further sliding. The process continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic cistemae resulting in masking of actin filaments and breakage of all cross-bridges. This cause the return of ‘Z’ lines along with filaments back to their original position, i.e., relaxation.

Q6. Differentiate between pectoral and pelvic girdle.
Ans: Pectoral and pelvic girdle help in the articulation of upper and lower limbs respectively. Each girdle is made of two equal halves. Each half of a pectoral girdle consists of clavicle and scapula. Scapula is a large triangular flat bone. There is glenoid cavity at the joint of scapula, clavicle and acromian process, which articulates with the head of humerus to form the shoulder joint. Each half of pelvic girdle is formed by three bones—ilium, ischium and pubis. At the point of their fusion; there is a cavity called acetabulum to which the head of femur articulates.

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NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-18-body-fluids-circulation/

NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation

Multiple Choice Questions

Q1. Which of the following cells does not exhibit phagocytic activity?
(a) Monocytes (b) Neutrophil (c) Basophil (d) Macrophage
Ans: (c) Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions.

Q2. One of the common symptoms observed in people infected with Dengue fever is
(a) significant decrease in RBC count
(b) significant decrease in WBC count
(c) significant decrease in platelets count
(d) significant increase in platelets count
Ans: (c) One of the common symptoms observed in people infected with Dengue fever is significant decrease in platelets count.

Q3. Which among the followings is correct during each cardiac cycle?
(a) The volume of blood pumped out by the Rt and Lt ventricles is same
(b) The volume of blood pumped out by the Rt and Lt ventricles is different
(c) The volume of blood received by each atrium is different
(d) The volume of blood received by the aorta and pulmonary artery is different
Ans: (a) The volume of blood pumped out by the Rt and Lt ventricles is same.

Q4. The cardiac activity could be moderated by the autonomous neural system. Tick the correct answer
(a) The parasympathetic system stimulates heart rate and stroke volume
(b) The sympathetic system stimulates heart rate and stroke volume
(c) The parasympathetic system decreases the heart rate but increase stroke volume
(d) The sympathetic system decreases the heart rate but increase stroke volume
Ans: (b) A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system (ANS). Neural signals through the sympathetic nerves (part of ANS) can increase the rate of heart beat, the strength of ventricular contraction and thereby the cardiac output. On the other hand, parasympathetic neural signals (another component of ANS) decrease the rate of heart beat, speed of conduction of action potential and thereby the cardiac output.

Q5. Mark the pair of substances among the following which is essential for coagulation of blood.
(a) Heparin and calcium ions (b) Calcium ions and platelet factors
(c) Oxalates and citrates (d) Platelet factors and heparin
Ans: (b) Calcium ions and platelet factors are essential for coagulation of blood.

Q6. ECG depicts the depolarisation and repolarisation processes during the cardiac cycle. In the ECG of a normal healthy individual one of the following waves is not represented.
(a) Depolarisation of atria (b) Repolarisation of atria
(c) Depolarisation of ventricles (d) Repolarisation of ventricles
Ans: (b) ECG depicts the depolarisation and repolarisation processes during the cardiac cycle. In the ECG of a normal healthy individual repolarisation of atria is not represented.

Q7. Which one of the following types of cells lack nucleus in humans
(a) RBC (b) Neutrophils (c) Eosinophils (d) Erythrocytes
Ans: (a and d) RBCs or erythrocytes lack nucleus in humans.

Q8. Which one of the following blood cells is involved in antibody production?
(a) B-lymphocytes (b) T-Lymphocytes
(c) RBC (d) Neutrophils
Ans: (a) B-lymphocytes cells are involved in antibody production.

Q9. The cardiac impulse is initiated and conducted further up to ventricle. The correct sequence of conduction of impulse is

Q10. Agranulocytes responsible for immune response of the body are
(a) Basophils (b) Neutrophils (c) Eosinophils (d) Lymphocytes
Ans: (d) Lymphocytes (20-25%) are of two major types—‘B’ and ‘T’ forms. Both B and T lymphocytes are responsible for immune responses of the body.

Q11. The second heart sound (dub) is associated with the closure of
(a) Tricuspid valve (b) Semilunar valves
(c) Bicuspid valve (d) Tricuspid and bicuspid valves
Ans: (b) The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves.

Q12. Which of the following correctly explains a phase/event in cardiac cycle in a standard electrocardiogram?
(a) QRS complex indicates atrial contraction
(b) QRS complex indicates ventricular contraction
(c) Time between S and T represents atrial systole
(d) P-wave indicates beginning of ventricular contraction
Ans: (b)
• P-wave—Atrial depolarisation (atrial systole/contraction)
• QRScomplex—Ventriculardepolarisation(ventricularsystole/contraction)
• T-wave—Ventricular repolarisation (ventricular relaxation)

Q13. Which of the following statements is incorrect?
(a) A person of ‘O’ blood group has anti ‘A’ and anti ‘B’ antibodies in his blood plasma
(b) A person of ‘B’ blood group cannot donate blood to a person of ‘A’ blood group
(c) Blood group is designated on the basis of the presence of antibodies in the blood plasma
(d) A person of AB blood group is universal recipient
Ans: (c) Blood group is designated on the basis of the antigen is present on the RBCs surface

Q14. What would be the cardiac output of a person having 72 heart beats per minute and a stroke volume of 50 mL?
(a) 360 mL (b) 3600 mL (c) 7200 mL (d) 5000 mL
Ans: (b) Cardiac output = Stroke volume x Heart rate = 50 x 72 = 3600 mL/min

Q15. Match the terms given under Column ‘A’ with their functions given under Column ‘B’ and select the answer from the options given below:

	Column A		Column B
A.	Lymphatic system	(i)	Carries oxygenated blood
B.	Pulmonary vein	(ii)	Immune Response
C.	Thrombocytes	(iii)	To drain back the tissue fluid to the circulatory system
D.	Lymphocytes	(iv)	Coagulation of blood

(a) A—(ii), B—(i), C—(iii), D—(iv)
(b) A—(iii), B—(i), C—(iv), D—(ii)
(c) A—(iii), B—(i), C—(ii), D—(iv)
(d) A—(ii), B—(i), C—(iii), D—(iv)

Ans: (b)

A.	Lymphatic System	(iii)	To drain back the tissue fluid to the circulatory system
B.	Pulmonary vein	(i)	Carries oxygenated blood
C.	Thrombocytes	(iv)	Coagulation of blood
D.	Lymphocytes	(ii)	Immune Response

 

Q16. Read the following statements and choose the correct option.
Statement 1: Atria receive blood from all parts of the body which subsequently flows to ventricles.
Statement 2: Action potential generated at sino-atrial node passes from atria to ventricles.
(a) Action mentioned in Statement 1 is dependent on action mentioned in Statement 2.
(b) Action mentioned in Statement 2 is dependent on action mentioned in Statement 1.
(c) Actions mentioned in Statements 1 and 2 are independent of each other.
(d) Actions mentioned in Statements 1 and 2 are synchronous.
Ans: (b) Statement 1: Atria receive blood from all pans of the body which subsequently flows to ventricles.
Statement 2: Action potential generated at sino-atrial node passes from atria to ventricles.
Action mentioned in Statement 2 is dependent on action mentioned in Statement 1.
Very Short Answer Type Questions
Q1. Name the blood component which is viscous and straw coloured fluid.
Ans: Plasma .

Q2. Complete the missing word in the statement given below:
a. Plasma without _______ factors is called serum.
b. _______ and monocytes are phagocytic cells.
c. Eosinophils are associated with _______ reactions.
d. _______ ions play a significant role in clotting.
e. One can determine the heart beat rate by counting the number of _______ in an ECG.

Ans: a. Plasma without clotting factors is called serum.
b. Neutrophils and monocytes are phagocytic cells.
c. Eosinophils are associated with allergic reactions.
d. Calcium ions play a significant role in clotting.
e. One can determine the heart beat rate by counting the number of QRS complex in an ECG.

Q3. Given below is the diagrammatic representation of a standard ECG. Label its different peaks

Q4. Name the vascular connection that exists between the digestive tract and liver.
Ans: Hepatic portal system

Q5. Given below are the abnormal conditions related to blood circulation. Name the disorders.
a. Acute chest pain due to failure of 0₂ supply to heart muscles
b. Increased systolic pressure
Ans: a. Acute chest pain due to failure of 0₂ supply to heart muscles—Angina
b. Increased systolic pressure—Hypertension/high blood pressure

Q6. Which is coronary artery diseases caused due to narrowing of the lumen of arteries?
Ans: Atherosclerosis

Q7. Define the following terms and give their-locations?
a. Purkinje fibre
b.Bundle of His
Ans: a. Purkinje fibre—Right and left bundles give rise to minute fibres throughout the ventricular musculature of the respective sides and are called purkinje fibres. .
b. Bundle of His—Purkinje fibres alongwith right and left bundles are known as bundle of His and present in ventricles.

Q8. State the functions of the following in blood:
a. Fibrinogen b. Globulin
c. Neutrophils d. Lymphocytes
Ans: a. Fibrinogen—Fibrinogens are needed for clotting or coagulation of blood. ■
b. Globulin—Globulins primarily are involved in immunity, i.e., defense mechanisms of the body.
c. Neutrophils—Phagocytosis
d. Lymphocytes—Immunity

Q9. What physiological circumstances lead to’erythroblastosis foetalis?
Ans: A special case of Rh incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus.

Q10. Explain the consequences of a situation in which blood does not coagulate.
Ans: This situation leads to excessive loss of blood from body due to injury which
can be fatal.
Q11. What is the significance of time gap in the passage of action potential from sino-atrial node to the ventricle?
Ans: This time gap is significant for ventricular systole.

Q12. How will you interpret an electrocardiogram (ECG) in which time taken in QRS complex is higher.
Ans: Period of ventricular systole increases

Short Answer Type Questions

Q1. The walls of ventricles are much thicker than atria. Explain.
Ans: The walls of ventricles are much thicker than atria because they pump blood more strongly than the atria.

Q2. Differentiate between:
a.  Blood and Lymph
b. Basophils and Eosinophils
c. Tricuspid and Bicuspid valves
Ans: a. Blood and Lymph: Blood is a connective tissue consisting of a fluid matrix, plasma and formed elements (RBCs, WBCs and Platelets). Blood flows in blood vascular system comprising heart, arteries and veins.
Lymph is a colourless fluid containing specialised lymphocytes (imparting immunity to the body), but devoid of RBCs. Lymph flows in the lymphatic system and it absorbs fats.

a.	Blood    ‘		Lymph
1.	It contains plasma, RBCs, WBCs and platelets	1.	It contains plasma and lymphocytes
2.	It is red in colour	2.	It is colourless
3.	Haemoglobin is present	3.	Haemoglobin is absent
4.	It transports nutrients and gases from heart to tissues and vice-versa	4.	It transports infection fighting white blood cells from tissues to lymph nodes
b.	Basophils		Eosinophils
1.	They constitute about 0.5-1% of WBCs	1.	They constitute about 2-3% of WBCs
2.	They secrete heparin, histamine and serotonin	2.	They resist infection
3.	They are involved in inflammatory reaction	3.	These are associated with allergic reactions
c.	Tricuspid valve		Bicuspid valve
1.	It has three cusps.	1.	It have two cusps.
2.	It is present between right atrium and right ventricle	2.	It is present between left atrium and left ventricle.

 

Q3. Briefly describe the following:
a. Anaemia
b. Angina Pectoris
c. Atherosclerosis
d. Hypertension
e. Heart failure
f. Erythroblastosis foetalis
Ans: a. Anaemia: Decrease in oxygen carrying capacity of blood either due to reduced RBCs production or low haemoglobin content is called anaemia.
b. Angina Pectoris: A symptom of acute chest pain appears when no enough oxygen is reaching the heart muscle. Angina can occur in men and women of any age but it is more common among the middle-aged ” and elderly. It occurs due to conditions that affect the blood flow.
c. Atherosclerosis: Sometimes deposition of calcium, fat, cholesterol and fibrous tissues occurs in the blood vessel (e.g., coronary artery) supplying blood to the heart muscles. This condition makes the lumen of arteries narrower affecting blood supply to heart; which leads to Coronary Artery Disease (CAD) also referred to as atherosclerosis.
d. Hypertension: If repeated checks of blood pressure of an individual is 140/90 (140 over 90) or higher, it shows hypertension. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.
e. Heart failure: Heart failure means the state of heart when it is not pumping blood effectively enough to meet the needs of the body. It is sometimes called congestive heart failure because congestion of the lungs is one of the main symptoms of this disease.
f. Erythroblastosis foetalis: A special case of Rh incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus! Rh antigens of the foetus do not get exposed to the Rh-negative blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However, during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh-positive blood from the foetus. In such cases, the mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh-negative) can leak
into the blood of the foetus (Rh-positive) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Q4. Explain the advantage of the complete partition of ventricle among birds and mammals and hence leading to double circulation.
Ans: Complete partition of ventricle among birds and mammals is advantageous because there is no mixing of oxygenated and deoxygenated blood in the ventricle, so tissues of the body receive more oxygenated blood.

Q5. What is the significance of hepatic portal system in the circulatory system?
Ans: The hepatic portal vein carries blood from intestine to liver before it is
delivered to systemic circulation. This is significant because excess of nutrients like glucose is converted into glycogen in liver and stored there.

Q6. Explain the functional significance of lymphatic system?
Ans: As the blood passes through the capillaries in tissues, some water along with many small water soluble substances move out into the spaces between the cells of tissues leaving the larger proteins and most of the formed elements in the blood vessels. This fluid released out is called the interstitial fluid or tissue fluid. It has the same mineral distribution as that in plasma. Exchange of nutrients, gases, etc., between the blood and the cells always occur through this, fluid. An elaborate network of vessels called the lymphatic system collects this fluid and drains it back to the major veins. The fluid present in the lymphatic system is called the lymph. Lymph is a colourless fluid containing specialised lymphocytes which are responsible for the immune responses of the body. Lymph is also an important carrier for nutrients, hormones, etc. Fats are absorbed through lymph in the lacteals present in the intestinal villi.

Q7. Write the features that distinguish between the two
a. Plasma and Serum
b. Open and Closed circulatory system
c. Sino-atrial node and Atrio-ventricular node

a.	Plasma		Serum
i.	Blood without              formed elements is called plasma	1.	Plasma without clotting factor is called serum
2.	Plasma has clotting factors	2.	Serum does not have clotting factors
3.	Plasma involved in blood coagulation	3.	Serum does not involve in blood coagulation
b.	Open circulatory system		Closed circulatory system
1.	Blood pumped by heart passes through large vessels into open spaces or body cavities called sinuses	1.	Blood pumped by the heart is always circulated through a closed network of blood vessels
2.	Less advantageous	2.	More advantageous
3.	Flow of fluid cannot be more precisely regulated	3.	Flow of fluid can be more precisely regulated
4.	It is present in arthropods molluscs, and hemichordates	4.	It is found in annelids and chordates
c.	Sino-atrial node		Atrio-ventricular node
1.	SA node is present in the right upper comer of the right atrium	1.	AV node is present in the lower left comer of the right atrium
2.	It initiates and maintains the rhythmic contractile activity of the heart	2.	It passes the electrical impulses from SA node to AV bundle
3.	It is also called pace-maker	3.	It is also called pace-setter

 

Q9. Answer the following:
a. Name the major site where RBCs are formed.
b. Which part of heart is responsible for initiating and maintaining its rhythmic activity?
c. What is specific in the heart of crocodiles among reptilians?
Ans: a. Bone marrow
b. Sino-Atrial Node (SA Node)
c. Reptiles have 3-chambered heart but crocodiles have 4-chambered heart.

Long Answer Type Questions
Q1. Explain Rh-incompatibility in humans.
Ans: A special case of Rh-incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh-negative blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However, during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh – positive blood from the foetus. In such cases, the mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh-negative) can leak into the blood of the foetus (Rh-positive) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Q2. Describe the events in cardiac cycle. Explain “double circulation”.
Ans:
• Cardiac cycle: To begin with, all the four chambers of heart are in a relaxed state, i.e., they are in joint diastole. As the tricuspid and bicuspid valves are open, blood from the pulmonary veins and vena cava flows into the left and the right ventricle respectively through the left and right atria. The semilunar valves are closed at this stage. The SAN now generates an action potential which stimulates both the atria to undergo a simultaneous contraction—the atrial systole. This increases the flow of blood into the ventricles by about 30%. The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of His transmits it through the entire ventricular musculature. This causes the ventricular muscles to contract (ventricular systole), the atria undergoes relaxation (diastole), coinciding with the ventricular systole. Ventricular systole increases the ventricular pressure causing the closure of tricuspid and bicuspid valves due to attempted backflow of blood into the atria. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery (right side) and the aorta (left side) are forced open, •allowing the blood in the ventricles to flow through these vessels into the circulatory pathways.
The ventricles now relax (ventricular diastole) and the ventricular pressure falls causing the closure of semilunar valves which prevents the backflow of blood into the ventricles. As the ventricular pressure declines further, the tricuspid and bicuspid valves are pushed open by the pressure in the atria exerted by the blood which was being emptied into them by the veins. The blood now once again moves freely to the ventricles. The ventricles and atria are now again in a relaxed (joint diastole) state, as earlier. Soon the SAN generates a new action potential and the events described above are repeated in that sequence and the process continues.
• Double circulation: The blood pumped by the right ventricle enters the pulmonary artery, whereas the left ventricle pumps blood into the aorta. The deoxygenated blood pumped into the pulmonary artery is passed on to the lungs from where the oxygenated blood is carried by the pulmonary veins into the left atrium. This pathway constitutes the pulmonary circulation. The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation. The systemic circulation provides nutrients, 02 and other essential substances to the tissues and takes C02 and other harmful substances away for elimination.

Q3. Explain different types of blood groups and donor compatibility by making a table.
Ans: ABO blood grouping is based* on the presence or absence of two surface antigens on the RBCs namely A and B. Similarly, the plasma of different individuals contain two natural antibodies anti-A and anti-B. Blood group ‘A’ carries antigen-A and antibodies-B. The donor’s group for blood group A are A and O. Blood group B carries antigen-B and antibodies-A. The donor’s group for blood group B are B and O. Blood group AB carries antigens A and B but no corresponding antibodies so, the compatible donor’s group for blood group AB are A, B, AB and O hence, blood group ‘AB’ is also called as “universal acceptor’’. Blood group ‘O’ carries no antigens but carries antibodies both A and B hence its compatible donor’s group is only ‘O’ but it is a compatible donor group for all the blood groups. A, B, AB and O hence, blood group ‘O’ is called as ‘universal donor’.

Blood groups and donor compatibility

Blood Group	. Antigen on RBCs	Antibodies in Plasma	Donor’s Compatibility
A	A	Anti-B	A, O
B	B	Anti-A	B, 0
AB	A. B	Nil	AB, A, B, 0
0 ‘	Nil	Anti – A, B	0

Q4. Write a short note on the following:
a. Hypertension
b. Coronary Artery Disease
Ans: a. Hypertension: If repeated checks of blood pressure of an individual is 140/90 (140 over 90) or higher, it shows hypertension. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.
b. Coronary Artery Disease: Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol and fibrous tissues, which makes the lumen of arteries narrower.

Q5. In the diagrammatic presentation of heart given below, mark and label, SAN, AVN, AV bundles, bundle of His and Purkinje fibres.

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NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-16-digestion-absorption/

NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption

Multiple Choice Questions

Q1. Select what is not true of intestinal villi among the following
(a) They possess microvilli
(b) They increase the surface area
(c) They are supplied with capillaries and the lacteal vessels (d_) They only participate in digestion of fats
Ans: (d) They only participate in digestion of fats.

Q2. Hepato-pancreatic duct opens into the duodenum and carries
(a) Bile (b) Pancreatic juice
(c) Both bile and pancreatic juice (d) Saliva
Ans: (c) Hepato-pancreatic duct opens into the duodenum and carries both bile and pancreatic juice.

Q3. One of the following is not a common disorder associated with digestive system
(a) Tetanus (b) Diarrhoea (c) Jaundice (d) Dysentery
Ans: (a) Tetanus is not a common disorder associated with digestive system.

Q4. A gland not associated with the alimentary canal is
(a) Pancreas (b) Adrenal
(c) Liver (d) Salivary glands
Ans: (b) A gland not associated with the alimentary canal is adrenal (this is endocrine gland).

Q5. Match the two columns and select the correct among options given

	Column I		Column II
A.	Biomacromolecules	(i)	Alimentary canal and of food associated gland
B.	Human digestive system	(ii)	Embedded in jawbones
C.	Stomach	(iii)	Outer wall of visceral organs

D.	Thecodont	(iv)	Converted into simple substances
E.	Serosa	(v)	J-shaped bag like structure

 

Options:
(a) A—(ii), B—(i), C—(v), D—(iii), E—(iv)
(b) A—(iv), B—(i), C—(v), D—(ii), E—(iii)
(c) A—(i), B—(ii), C—(iii), D—(iv), E—(v)
(d) A—(i), B—(iii), C—(ii), D—(iv), E-(v)

Ans. (b)

A.	Biomacromolecules	(iv)	Converted into simple substances
B.	Human digestive	(i)	Alimentary canal and of food associated
	system		gland
C.	Stomach	(v)	J-shaped bag like structure
D.	Thecodont	(ii)	Embedded in jawbones
E.	Serosa	(iii)	Outer wall of visceral organs

6. Match the two columns and select the right one among options given

Column I		Column II
A.	Duodenum	(i)	A cartilagenous flap
B.	Epiglottis	(ii)	Small blind sac
C.	Glottis	(Hi)	‘C’ shaped structure emerging from the stomach
D.	Caecum	(iv)	Opening of wind pipe

(a) A—(i), B—(ii), C—(iii), D—(iv)
(b) A—(iv), B—(iii), C—(ii), D—(i)
(c) A—(iii), B—(i), C—(iv), D—(ii)
(d) A—(ii), B—(iv), C—(i), D—(iii)

A.	Duodenum	(iii)	‘C’ shaped structure emerging from the stomach
B.	Epiglottis	(i)	A cartilagenous flap
C.	Glottis	(iv)	Opening of wind pipe
D.	Caecum	(ii)	Small blind sac

Q7. Match the enzymes with their respective substrates choose the right one among options given.

Column I		Column II
A.	Lipase	(i)	Dipeptides
B.	Nuclease	(ii)	Fats
C.	Carboxypeptidase	(iii)	Nucleic acids
D.	Dipeptidases	(iv)	Proteins, peptones and proteoses

Options:
(a) A—(ii), B—(iii), C—(i), D—(iv)
(b) A—(iii), B—(iv), C—(ii), D—(i)
(c) A—(iii), B—(i), C—(iv), D—(ii)
(d) A—(ii), B—(iii), C—(iv), D—(i)

 

A.	Lipase	(ii)	Fats
B.	Nuclease	(iii)	Nucleic acids
C.	Carboxypeptidase	(iv)	Proteins, peptones and proteoses
D.	Dipeptidases	(i)	Dipeptides

Ans. (d)

Q9. Liver is the largest gland and is associated with functions, choose one which is not correct
(a) Metabolism of carbohydrate
(b) Digestion of fat
(c) Formation of bile
(d) Secretion of hormone called gastrin
Ans: (d) Liver is the largest gland and is associated with functions metabolism of carbohydrate, digestion of fat and formation of bile.

Q10. Mark the right statement among the following
(a) Trypsinogen is an inactive enzyme
(b) ’ Trypsinogen is secreted by intestinal mucosa
(c) Enterokinase is secreted by pancreas
(d) Bile contains trypsin
Ans: (a)

Very Short Answer Type Questions

Q1. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall. What do we call the food then?
Ans: Chyme

Q2. Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase, activates it. Which tissue/cells secrete this enzyme?/ How is it activated?
Ans: Intestinal mucosa

Q3. In which part of alimentary canal does absorption of water, simple sugars and alcohol takes place?
Ans: Stomach

Q4. Name the enzymes involved in the breakdown of nucleotides into sugars and bases.
Ans: Nucleotidases and Nucleosidases

Q5. Define digestion in one sentence.
Ans: The process of conversion of complex food substances in the digestive system to simple absorbable forms is called digestion.

Q6. What do we call the type of teeth attachment to j aw bones in which each tooth is embedded in a socket of jaws bones?
Ans: Thecodont

Q7. Stomach is located in upper left portion of the abdominal cavity and has three major parts. Name these three parts.
Ans: Cardiac, fundic and pyloric

Q8. Does gall bladder make bile?
Ans: No, it only stores.

Q9. Correct the following statements by deleting one of entries (given in bold).
a. Goblet cells are located in the intestinal mucosal epithelium and secrete chymotrypsin/mucus.
b. Fats are broken down into di- and monoglycerides with the help of amylase/lipases.
c. Gastric glands of stomach mucosa have oxyntic cell/chief cells which secrete HCl.
d. Saliva contains enzymes that digest starch/protein.
Ans: a. Goblet cells are located in the intestinal mucosal epithelium and secrete _ mucus.
b. Fats are broken down into di- and monoglycerides with the help of lipases.
c. Gastric glands of stomach mucosa have oxyntic cell which secrete HC1.
d. Saliva contains enzymes that digest starch.

Short Answer Type Questions
Q1. What is pancreas? Mention the major secretions of pancreas that are helpful in digestion.
Ans: Pancreas is a gland having exocrine and endocrine portions involved in secreting digestive enzymes as well as hormones. Major secretions of pancreas involved in digestion are inactive enzymes listed below:
a. Trypsinogen
b. Chymotrypsinogen
c. Procarboxypeptidases
d.Amylases
e. Lipases
f. Nucleases

Q2. Name the part of the alimentary canal where major absorption of digested food takes place. What are the absorbed forms of different kinds of food materials? ‘
Ans: Small intestine is the part of alimentary canal where digested food materials are mainly absorbed. Amino acids (proteins), monosaccharides like glucose, fructose, galactose, etc. (carbohydrate) and fatty acids and glycerol (fats) are different absorbable forms of food materials.

Q3. List the organs of human alimentary canal and name the major digestive glands with their location.
Ans: The alimentary canal begins with an anterior opening the mouth and then buccal cavity, pharynx, oesophagus, stomach, small intestine, large intestine and it opens out posteriorly through the anus.
The digestive glands associated with the alimentary canal include the salivary glands, the liver and the pancreas. Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub¬mandibular (lower jaw) and the sublinguals (below the tongue). Liver is situated in the abdominal cavity, just below the diaphragm. The pancreas is situated between the limbs of the ‘C’-shaped duodenum.

Q4. What is the role of gall bladder? What may happen if it stops functioning or is removed?
Ans: The bile secreted by the hepatic cells passes through the hepatic ducts and is stored and concentrated in a thin muscular sac called the gall bladder. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases. If it stops functioning or is removed then digestion of fat will be affected.

Q5. Correct the statement given below by the right option shown in the bracket against them.
a. Absorption of amino acids and glycerol takes place in the (small intestine/ large intestine).
b. ‘The faeces in the rectum initiate a reflex causing an urge for its removal, (neural /hormonal)
c. Skin and eyes turn yellow in infection, (liver/stomach)
d. Rennin is a proteolytic enzyme found in gastric juice in (infants/adults).
e. Pancreatic juice and bile are released through (intestinepancreatic/ hepato-pancreatic duct).
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in region of small intestine, (jejunum/duodenum)
Ans: a. Absorption of amino acids and glycerol takes place in the small intestine.
b. The faeces in the rectum initiate a neural reflex causing an urge for its removal.
c. Skin and eyes turn yellow in infection of liver. .
d. Rennin is a proteolytic enzyme found in gastric juice in infants.
e. Pancreatic juice and bile are released through hepato-pancreatic duct.
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in duodenum region of small intestine.

Q6. What are three major types of cells found in the gastric glands? Name their secretions.
Ans: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).
Q7. How is the intestinal mucosa’protectcd from the acidic food entering from stomach?
Ans: The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from excoriation by the highly concentrated hydrochloric acid.

Q8. How are the activities of gastro-intestinal tract regulated?
Ans: The activities of the gastro-intestinal tract are under neural and hormonal control for proper coordination of different parts. The sight, smell and/or the presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also, similarly, stimulated by neural signals. The muscular activities of different parts of the alimentary canal can also be moderated by neural mechanisms, both local and through CNS. Hormonal control of the secretion of digestive juices is carried out by local hormones produced by the gastric and intestinal mucosa.

Q9. Distinguish between constipation and indigestion. Mention their major causes.
Ans: Constipation: In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly.
Indigestion: In this condition, the food is not properly digested leading to a feeling of fullness. The causes of indigestion are inadequate enzyme secretion, anxiety, food poisoning, over eating, and spicy food.

10. Describe the enzymatic action on fats in the duodenum.
Ans: In the duodenum fats are broken down by pancreatic lipases with the help of bile into di- and monoglycerides.

Long Answer Type Questions
Q1. A person had roti and dal for his lunch. Trace the changes in those during its passage through the alimentary canal.
Ans: The process of digestion is accomplished by mechanical and chemical processes. The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About 30% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide – maltose.
• The stomach stores the food for 4-5 hours. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme. The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).
• The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
Proteins proteoses and peptones (partially hydrolysed proteins) in the chyme reaching the intestine are acted upon by the proteolytic enzymes of pancretic juice as given below.

Q2. What are the various enzymatic types of glandular secretions in our gut helping digestion of food? What is the nature of end products obtained after complete digestion of food?
Ans: Enzymatic types of glandular secretions in our gut:
a. Salivary glands: Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub-mandibular (lower jaw) and the sublinguals (below the tongue). These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.
b. Gastric glands: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).
c. The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes.

Q3. Discuss mechanisms of absorption.
Ans: Mechanisms of absorption: Absorption is the process by which the end products of digestion pass through the intestinal mucosa into the blood or lymph. It is carried out by passive, active or facilitated transport mechanisms. Small amounts of monosaccharides like glucose, amino acids and some electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon the concentration gradients. However, some substances like glucose and amino acids are absorbed with the help of carrier proteins. This mechanism is called the facilitated transport.
Transport of water depends upon the osmotic gradient. Active transport occurs against the concentration gradient and hence requires energy. Various nutrients like amino acids, monosaccharides like glucose, electrolytes like Na+ are absorbed into the blood by this mechanism.

Q4. Discuss the role of hepato-pancrdatic complex in digestion of carbohydrate, protein and fat components of food.
Ans: The bile, pancreatic juice and the intestinal juice are the secretions released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes— trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases.
Proteins proteoses and peptones (partically hydrolysed proteins) in the chime reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice as given below:

Q5. Explain the process of digestion in the buccal cavity with a note on the arrangement of teeth.
Ans: The process of digestion in the buccal cavity: The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide—maltose. Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

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NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-15-plant-growth-development/

NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development

Multiple Choice Questions

Q1. Ethylene is used for
(a) Retarding ripening of tomatoes
(b) Hastening of ripening of fruits
(c) Slowing down ripening of apples
(d) Both (b) and (c)
Ans: (b) Ethylene is used for hastening of ripening of fruits.

Q2. Coconut water contains
(a) ABA
(b) auxin
(c) cytokinin
(d) gibberellin
Ans: (c) Coconut water contains cytokinin.

Q3. The effect of apical dominance can be overcome by which of the following hormone?
(a) IAA
(b) Ethylene
(c) Gibberellin
(d) Cytokinin
Ans: (d) The effect of apical dominance can be overcome by cytokinin hormone.

Q4. Match the following:

A.	IAA	(0	Herring sperm DNA
B.	ABA	(ii)	Bolting
C.	Ethylene	(iii)	Stomatal closure
D.	GA	(iv)	Weed-free lawns
E.	Cytokinins	(v)	Ripening of fruits

Ans: (a)

A.	IAA	(iv)	Weed-free lawns
B.	ABA	(iii)	Stomatal closure
C.	Ethylene	(v)	Ripening of fruits
D.	GA	(ii)	Bolting
E.	Cytokinins	(i)	Herring sperm DNA

 

Q5. Apples are generally wrapped in waxed paper to
(a) prevent sunlight for changing its colour
(b) prevent aerobic respiration by checking the entry of 02
(c) prevent ethylene formation due to injury
(d) make the apples look attractive
Ans: (b) Apples are generally wrapped in waxed paper to prevent aerobic respiration by checking the entry of 02.

Q6. Growth can be measured in various ways. Which of these can be used as parameters to measure growth?
(a) Increase in cell number
(b) Increase in cell size
(c) Increase in length and weight
(d) All the above
Ans: (d) Growth can be measured in various ways. Increase in cell number, increase in cell size and increase in length and weight are used as parameters to measure growth.

Q7. The term synergistic action of hormones refers to
(a) when two hormones act together but bring about opposite effects
(b) when two hormones act together and contribute to the same function
(c) when one hormone affects more than one function
(d) when many hormones bring about any one function
Ans: (b) The term synergistic action of hormones refers to when two hormones act together and contribute to the same function.

Q8. Plasticity in plant growth means that
(a) plant roots are extensible
(b) plant development is dependent on the environment
(c) stems can extend
(d) none of the above
Ans: (b) Plasticity in plant growth means that plant development is dependent on the environment.

Q9. To increase sugar production in sugarcanes, they are sprayed with
(a) IAA
(b) cytokinin
(c) gibberellin
(d) ethylene
Ans: (c) To increase sugar production in sugarcanes, they are sprayed with gibberellin.

Q10. ABA acts antagonistic to
(a) ethylene
(b) cytokinin
(c) gibberellic acid
(d) IAA
Ans: (c) ABA acts antagonistic to gibberellic acid.

Q11. Monocarpic plants are those which
(a) bear flowers with one ovary
(b) flower once and die
(c) bear only one flower
(d) all of the above
Ans: (b) Monocarpic plants are flower once and die.

Q12. The photoperiod in plants is perceived at
(a) meristem
(b) flower
(c) floral buds
(d) leaves
Ans: (d) The photoperiod in plants is perceived at leaves.

Very Short Answer type Questions
Q1. Fill in the places with appropriate word/words.
a. A phase of growth which is maximum and fastest is .
b. Apical dominance as expressed in dicotyledonous plants is due to the
presence of more _____ in the apical bud than in the lateral ones.
c. In addition to auxin, a ________ must be supplied to culture medium to
obtain a good callus in plant tissue culture.
d. _________of a vegetative plants are the sites of photoperiodic perception.
Ans: a. Exponential/log phase of an S-curve.
b. Auxin/IAA
c. CytokininlKinetinl6 BAP/Zeatinletc.
d. Leaves

Q2. PJ.ant growth substances (PGS) have innumerable practical applications. Name the PGS you should use to
a. increase yield of sugar cane b. promote lateral shoot growth
c. cause sprouting of potato tuber d. inhibit seed germination
Ans: a. GA3/gibberellinlgibberellic acid
b. CytokininlzeatinlkinetinJKn
c. C2H4/Ethylene
d. ABA/Abscisic acid

Q3. A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.

Q4. Gibberellins were first discovered in Japan when rice plants were suffering from bakane (the foolish seedling disease) caused by a fungus Gibberella fujikuroi.
a. Give two functions of this priytohormone.
b. Which property of Gibberellin caused foolish seedling disease in rice?
Ans: a. GA3 is used to speed up the malting process in brewing industry. Gibberellins also promote bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.
b. Gibberellin causes foolish seedling disease in rice because it has the property of internode elongation.

Q5. Gibberellins promote the formation of _________ flowers on genetically _________ plants in Cannabis whereas ethylene promotes formation of _______ flowers
on genetically ____ plants.
Ans: Gibberellins promote the formation of male flowers on genetically female plants in Cannabis whereas ethylene promotes formation of female flowers on genetically male plants.

Q6. Classify the following plants into Long-Day Plants (LDP), Short-Day Plants (SDP) and Day-Neutral Plants (DNP) Xanthium, Henbane (Hyoscyamus niger), Spinach, Rice, Strawberry, Bryophyllum, Sunflower, Tomato, Maize.
Ans: Xanthium: (SDP)
Henbane (Hyoscyamus niger): (LDP)
Spinach: (LDP)
Rice: (SDP)
Strawberry: (SDP)
Bryophyllum: LSDP (Long short day plants)
Sunflower: (DNP)
Tomato: (DNP)
Maize: (DNP)

Q7. A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which can plant growth regulator be applied to achieve this?
Ans: Ethylene (C2H4)

Q8. Where are the following hormones synthesised in plants?
a. IAA
b. Gibberellins
c. Cytokinins
Ans: a. IAA: Shoot tips and apical bud
b. Gibberellins: Root tips and young leaves
c. Cytokinins: Meristematic zones like root tips

Q9. In botanical gardens and tea gardens, gardeners trim the plants regularly so that they remain bushy. Does this practice have any scientific explanation?
Ans: Mostly in higher plants, the growing apical bud inhibits the growth of the lateral (axillary) buds, a phenomenon called apical dominance. Removal of shoot tips (decapitation) usually results in the growth of lateral buds. Hence, in botanical gardens and tea gardens, gardeners trim the plants regularly so that they remain bushy.

Q10. Light plays an important role ‘in the life of all organisms. Name any three physiological processes in plants which are affected by light.
Ans: Photoperiodism, phototropism and photosynthesis.

Q11. In the figure of Sigmoid growth curve given below, label segments 1,2 and 3

Q12. Growth is one of the characteristics of all living organisms. Do unicellular organism also grow? If so, what are the parameters?
Ans: Increase in mass and increase in number of individuals are twin characteristics of growth. A multicellular organism grows by cell division. Unicellular organisms grow by cell division. One can easily observe this in in vitro cultures by simply counting the number of cells under the microscope.

Q13. The rice seedlings infected with fungus Gibberella fujikuroi is called foolish seedlings? What was the reason behind it?
Ans: The rice seedling infected with fungus Gibberella fujikuroi is called foolish seedlings because the fungus secreted a hormone gibberellin and causes excessive growth of rice plants. Plants become tall but unable to produce seeds so they are called foolish.

Short Answer Type Questions

Q1. Nicotiana tobacum, a short-day plant, when exposed to more than critical period of light fails to flower. Explain.
Ans: a. Some plants require a periodic exposure to alternate light and dark for its flowering response. This phenomenon is termed photoperiodism.
b. The requirement of light exposure is critical. The SDP plants, when exposed to light period in excess of critical period fail to flower,
c. Those plants which require exposure to light period at critical or more than critical period for its flowering response are called long-dayplant.
d. Nicotiana tabacum fails to flower if exposed to more than critical period of light because it is an SDP.

Q2. What are the structural characteristics of
a. Meristematic cells near root tip
b. The cells in the elongation zone of the root
Ans: a. The meristematic cells near root tip are characterised by
• rich protoplasm
• large conspicuous nucleus
• thin and cellulosic cell wall -primary in nature
• fewer vacuoles
• greater number of mitochondria
• numerous (abundant) plasmodesmata
b. The cells in the elongation zone of a root are characterized by
• increased vacuolation
• enlarged size/dimension
• deposition of new cellulosic cell walls

Q3. Does the growth pattern in plants differ from that in animals? Do all the parts of plant grow indefinitely? If not, name the regions of plant, which can grow indefinitely.
Ans: Yes, the growth pattern in plants differ from that in animals. Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

Q4. Explain in 2-3 lines each of the following terms with the help of examples taken from different plant tissues.
a. Differentiation
b. De-differentiation
c. Re-differentiation
Ans: a. Differentiation: The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation. During differentiation, cells undergo few to major structural changes both in their cell walls and protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm. They also develop a very strong, elastic, lignocellulosic secondary cell walls, to carry water to long distances even under extreme tension.
b. De-differentiation: The living differentiated cells that by now have lost the capacity to divide can regain the capacity of division under certain conditions. This phenomenon is termed as de-differentiation. For example, formation of meristems – interfascicular cambium and cork cambium from fully differentiated parenchyma cells.
c. Re-differentiation: While doing de-differentiation, such meristems/ tissues are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions, i.e., get re-differentiated, e.g., secondary xylem and secondary cortex.

Q5. Auxins are growth hormones capable of promoting cell elongation. They have been used in horticulture to promote growth, flowering and rooting. Write a line to explain the meaning of the following terms related to auxins.
a. Auxin precursors
b. Anti-auxins
c. Synthetic auxins
Ans: a. Auxin precursors: The substances that produce the auxin are called auxin precursors. For example, tryptophan is the auxin precursor.
b. Anti-auxins: The substances which inhibit the synthesis or transport of auxin are called anti-auxins. For example, TIBA (Triiodobenzoic acid) is anti-auxin compound.
c. Synthetic auxins: The artificially synthesised chemicals having auxin-like property are called synthetic auxins. For example, NAA
– (Naphthalene acetic acid) and 2, 4-D (2, 4-Dichloro phenoxyacetic acid).

Q6. The role of ethylene and abscisic acid is both positive and negative. Justify the statement.
Ans: Positive roles of ethylene: Influences of ethylene on plants include horizontal growth of seedlings, swelling of the axis and apical hook formation in dicot seedlings. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds, sprouting of potato tubers.
• Negative roles of ethylene: Ethylene promotes senescence and abscission of plant organs especially of leaves and flowers.
• Positive roles of abscisic acid: ABA plays an important role in seed development and maturation.
• Negative roles of abscisic acid: It acts as a general plant growth inhibitor and an inhibitor of plant metabolism. ABA inhibits seed germination. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Q7. While experimentation, why do you think it is difficult to assign any effect seen to any single hormone?
Ans: Many hormones have synergistic and antagonistic effect with each other. So, while experimentation, it is difficult to assign any effect seen to any single hormone.

Q8. What is the mechanism underlying the phenomenon by which the terminal/ apical bud suppresses the growth of lateral buds? Suggest measures to overcome this phenomenon.
Ans: The phenomenon by which the terminal/apical bud suppresses the growth of lateral buds is called apical dominance. Apical dominance is due to auxin hormone secreted by apical buds. This can be overcome by decapitation (removal of apical buds) or the application of cytokinin.

Q9. In animals there are special glands secreting hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?
Ans: In plants, the hormones are formed by different tissues like shoot tips, root tips, meristematic tissues, leaves and apical buds, etc.
Hormones are translocated to the site of activity by vascular tissues (xylem and phloem)

Q10. Many discoveries in science have been accidental. This is true for plant hormones also. Can you justify this statement by giving an example? Also what term is used for such accidental findings?
Ans: The discovery of each of the five major groups of PGRs have been accidental. All this started with the observation of Charles Darwin and his son Francis Darwin when they observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source (phototropism). After a series of experiments, it was concluded that the tip of coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from tips of coleoptiles of oat seedlings. Such accidental findings or discoveries are known as serendipity.

Q11. To get carpet-like grass lawn are mowed regularly. Is there any scientific explanation for this?
Ans: To get a carpet-like grass lawns are mowed regularly because mowing causes decapitation which promotes the growth of lateral buds.

Q12. In a slide showing different types of cells, can you identify which type of the cell may be meristematic and the one which is incapable of dividing and how?
Ans: The meristematic cells are rich in protoplasm, possess large conspicuous nuclei. Their cell walls are primary in nature, thin and cellulosic with abundant plasmodesmatal connections. Cells incapable of dividing attain their maximal size in terms of wall thickening and protoplasmic modifications.

Q13. A rubber band stretches and reverts back to its original position. Bubble gum stretches, but it would not return to its original position. Is there any difference between the two processes? Discuss it with respect to plant growth (Hint: Elasticity (reversible), Plasticity (irreversible))
Ans: A rubber band stretches and reverts back to its original position, it is due to elasticity. Bubble gum stretches, but it would not return to its original position, this is due to plasticity.
• The meristematic cells are rich in protoplasm, possess large conspicuous nuclei. Their cell walls are primary in nature, thin, cellulosic and elastic with abundant plasmodesmatal connections.
• Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
e. g., heterophylly in cotton, coriander and larkspur. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.

Q14. Label the diagram
a. This is which part of a dicotyledonous plant?
b. If we remove part 1 from the plant, what will happen?

Q15. Both animals and plants grow. Why do we say that growth and differentiation in plants is open and not so in animals? Does this statement hold true for sponges also?
Ans: Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the
Plant Growtli and Development 167
activity of the meristem is called the open form of growth. Yes, this statement hold true for sponges also.

Q16. Define parthenocarpy. Name the plant hormone used to induce parthenocarpy.
Ans: Most fruits however develop only from the ovary and are called true fruits.
Although in most of the species, fruits are the results of fertilisation, there are a few species in which fruits develop without fertilisation. Such fruits are called parthenocarpic fruits. Banana is one such example. Parthenocarpy can be induced through the application of growth hormones (like gibberellin and auxin) and such fruits are seedless. Auxins induce parthenocarpy in tomatoes.

Q17. While eating watermelons, all of us wish it was seedless. As a plant physiologist can you suggest any method by which this can be achieved.
Ans: This can be achieved through parthenocarpy. Parthenocarpy can be induced through the application of growth hormones (like gibberellin and auxin) and such fruits are seedless.

Q18. A gardener finds some broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently?
Ans: The dicotyledonous plants grow by their apical shoot meristems while grasses (which make lawns) possess intercalary meristem. Certain auxins, such as synthetic 2, 4-Dichlorophenoxyacetic acid (2,4-D) when applied in excess can damage the shoot apical meristems but they do not cause any damage to the- intercalary meristems. Thus, when 2, 4-D is sprayed on lawns, only the dicots get killed and the lawns become free of weeds.

Q19. On germination a seed first produces shoots with leaves, flowers appear later,
a. Why do you think this happens?
b. How is this advantageous to the plant?
Ans: a. All organisms have to reach a certain stage of growth and maturity in their life, before they can reproduce sexually. That period of growth is called the juvenile phase. It is known as vegetative phase in plants. This phase is of variable durations in different organisms. The end of juvenile/ vegetative phase which marks the beginning of the reproductive phase can be seen easily in the higher plants when they come to flower.
b. This enables the plant to have sufficient time to reach maturity.

Q20. Fill in the blanks:
a. Maximum growth is observed in phase.
b. Apical dominance is due to .
c. initiate rooting. .
d. Pigment involved in Photoperception in flowering plants is .
Ans: a. Maximum growth is observed in log/exponential phase.
b. Apical dominance is due to auxin.
c. Auxins initiate rooting.
d. Pigment involved in Photoperception in flowering plants is
phytochrome.

Long Answer Type Questions ‘
Q1. Some varieties of wheat are known as spring wheat while others are called winter wheat. Former variety is sown, and planted in spring and is harvested by the end of the same season. However, winter varieties, if planted in spring, fail to flower or produce mature grains within a span of a flowering season. Explain, why?
Ans: There are plants for which flowering is either quantitatively or qualitatively dependent on exposure to low temperature. This phenomenon is termed vernalisation. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity. Vernalisation refers specially to the promotion of flowering by a period of low temperature. Some important food plants, wheat, barley, rye have two kinds of varieties: winter and spring varieties. The ‘spring’ variety are normally planted in the spring and come to flower and produce grain before the end of the growing season. Winter varieties, however, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season. Hence, they are planted in autumn. They germinate, and over winter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.

Q2. It is known that some varieties of wheat are sown in autumn but are harvested around next mid-summer.
a. ■ What could be the probable reason for this?
b. What term is used for this promotion of flowering under low temperature?
c. Which plant hormone can replace the cold treatment?
Ans: a. Winter varieties, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season. Hence, they are planted in autumn. They germinate, and over winter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.
b. Vernalisation
c. Gibberellin

Q3. Name a hormone which
a. is gaseous in nature
b. is responsible for phototropism
c. induces femaleness in flowers of cucumber
d. is used for killing weeds (dicots)
e. induces flowering in long day plants
Ans: a. Gaseous in nature: Ethylene (C2H4)
b. Responsible for phototropism: Auxin
c. Induces femaleness in flowers of cucumber: Ethylene (C2H4)
d. Used for killing weeds (dicots): Auxin
e. Induces flowering in long day plants: Gibberellin

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NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

Multiple Choice Questions

Q1. The ultimate electron acceptor of respiration in an aerobic organism is
(a) Cytochrome (b) Oxygen
(c) Hydrogen (d) Glucose
Ans: (b) The ultimate electron acceptor of respiration in an aerobic organism is oxygen.

Q2. Phosphorylation of glucose during glycolysis is catalyzed by
(a) Phosphoglucomutase (b) Phosphoglucoisomerase
(c) Hexokinase (d) Phosphorylase
Ans: (c) Phosphorylation of glucose during glycolysis is catalyzed by hexokinase.

Q3. Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms
(a) Lactic acid
(b) C02 + H20
(c) Acetyl CoA + C02
(d) Ethanol + CO
Ans: (c) Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms Acetyl CoA + CO2.

Q4. Electron Transport System (ETS) is located in mitochondrial
(a) Outer membrane
(b) Inter membrane space
(c) Inner membrane
(d) Matrix
Ans: (c) Electron Transport System (ETS) is located in mitochondrial inner membrane.

Q5. Which of the following exhibits the highest rate of respiration? .
(a) Growing shoot apex
(b) Germinating seed
(c) Root tip
(d) Leaf bud
Ans: (b) Germinating seed exhibits the highest rate of respiration.

Q6. Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria synthesise ATP
(b) Mitochondria have a double membrane .
(c) The en2ymes of the Krebs’ cycle and the cytochromes are found in mitochondria
(d) Mitochondria are found in almost all plant and animal cells
Ans: (a) Mitochondria are called powerhouses of the cell because mitochondria synthesise ATP.

Q7. The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP + H20
Ans: (d) The end product of oxidative phosphorylation isATP+H₂o.

Q8. Match the following and choose the correct option from those given below.

Column A		Column B
A.	Molecular oxygen	(i)	a-Ketoglutaric acid
B.	Electron acceptor	(ii)	Hydrogen acceptor
C.	Pyruvate dehydrogenase	(iii)	Cytochrome C
D.	Decarboxylation	(iv)	Acetyl Co A

(a) A—(ii), B—(iii), C—(iv), D—(i)
(b) A—(iii), B—(iv), C—(ii), D—(i)
(c) A—(ii), B—(i), C—(iii), D—(iv)
(d) A—(iv), B—(iii), C—(i), D—(ii)

Ans. (a)

A.	Molecular oxygen	(ii)	Hydrogen acceptor
B.	Electron acceptor	(iii)	Cytochrome C
C.	Pyruvate dehydrogenase	(iv)	Acetyl Co A
D.	Decarboxylation	(i)	a-Ketoglutaric acid

Very Short Answer Type Questions

Q1. Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?
Ans: Energy released during the oxidation of compounds in respiration is immediately stored in ATP in the form of chemical bonds.
ADP + iP + energy —> ATP
As and when needed, this bond energy is broken and utilised
ATP —> ADP + iP + energy

Q2. Explain the term “Energy Currency”. Which substance acts as energy currency in plants and animals?
Ans: Every function of the cell requires energy. Energy currency stores and releases the energy as and when needed in the cell. ATP is called energy currency in both plants and animals.

Q3. Different substrates get oxidized during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?

Q5. When does anaerobic respiration occur in man and yeast?
Ans: In alcoholic fermentation (by yeast), incomplete oxidation of glucose occurs under anaerobic conditions by sets of reactions where pyruvic acid is converted to ethanol and C02.
PA —> Ethanol + C02
In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid.

Q6. Which of the following will release more energy on oxidation? Arrange them in ascending order?
a. 1 g of fat b. 1 g of protein
c. 1 g of glucose
d. 0.5 g of protein + 0.5 g glucose

Q7. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively ________ and ______.
Ans. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively pyruvic acid and ethanol + CO₂.

Short Answer Type Questions

Q1. If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.
Ans: Glucose or fruit juice is absorbed easily through the alimentary canal. In the cells glucose is oxidised and energy is released immediately. A cheese sandwich provides energy after digestion and absorption which takes long time.

Q2. What is meant by the statement “aerobic respiration is more efficient”?
Ans: Aerobic respiration is more efficient because fermentation accounts for
only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to C02 and H20. Also, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q3. Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions? Write in the space provided in the diagram.

Q4. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms?
Ans: Anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms due to the oxygen deficiency. In our skeletal muscles during strenuous exercise oxygen deficiency leads to anaerobic respiration. In germinating seeds anaerobic respiration leads to release of energy for emerging the seedling from the soil.

Q5. Oxygen is an essential requirement for aerobic respiration but it enters the respiratory process at the end? Discuss.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor.

Q6. Respiration is an energy releasing and enzymatically controlled catabolic process which involves a step-wise oxidative breakdown of organic substances inside living cells. In this statement about respiration explain the meaning of
1. Step-wise oxidative breakdown and
2. Organic substances (used as substrates).
Ans: 1. Step-wise oxidative breakdown: During oxidation within a cell, all the energy contained in respiratory substrates is not released free into
the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP. Hence, it is important to understand that the energy released by oxidation in respiration is not (or rather cannot be) used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised.
2. Organic substances (used as substrates): The compounds that are oxidised during this process are known as respiratory substrates. Usually carbohydrates are oxidised to release energy, but proteins, fats and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Q7. Comment on the statement – Respiration is an energy producing process but ATP is being used in some steps of the process.
Ans: In the respiration pathway, there are some steps where energy is utilised for phosphorylation. For example, conversion of glucose to glucose-6-phosphate consume one ATP. But at the end of the respiratory process many more ATP are produced.

Q8. The figure given below shows the steps in glycolysis. Fill in the missing steps A, B, C, D and also indicate whether ATP is being used up or released at step E?

Q9. Why is respiratory pathway referred to as an amphibolic pathway? Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Q10. We commonly call ATP as the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.
Ans: Yes, some other energy carriers are also present in a cell like GTP (guanosine triphosphate), ADP (adenosine diphosphate) and creatine phosphate.

Q11. ATP produced during glycolysis is a result of substrate level phosphorylation. Explain.
Ans: ATP produced during glycolysis is a result of substrate level phosphorylation because these ATP are produced without the electron transport system (ETS) and chemiosmosis.
During substrate level phosphorylation ATP is directly synthesised from ADP and inorganic phosphate (iP).

Q12. Do you know any step in the TCA cycle where there is substrate level phosphorylation. Which one?
Ans: During TCA cycle there is one step where substrate level phosphorylation takes place. This occurs during conversion of succinyl-CoA to succinic acid.

Q13. In a way green plants and cyanobacteria have synthesised all the food on the earth. Comment.
Ans: All the energy required for ‘life’ processes is obtained by oxidation of some macromolecules that we call ‘food’. Only green plants and cyanobacteria can prepare their own food; by the process of photosynthesis they trap light energy and convert it into chemical energy that is stored in the bonds of carbohydrates like glucose, sucrose and starch.

Q14.When a substrate is being metabolised, why does not all the energy that is produced get released in one step? It is released in multiple steps. What is the advantage of step-wise release?
Ans: The complete combustion of glucose, which produces C0₂ and H₂0 as end products, yields energy most of which is given out as heat.
C₆H₁₂O₆  +  6O₂  —> 6CO₂+ 6H₂O + Energy
If this energy is to be useful to the cell, it should be able to utilise it to synthesise other molecules that the cell requires. The strategy that the plant cell uses is to catabolise the glucose molecule in such a way that not all the liberated energy goes out as heat. The key is to oxidise glucose not in one step but. in several small steps enabling some steps to be just large enough such that the energy released can be coupled to ATP synthesis.

Q15. Respiration requires 02. How did the first cells on the earth manage to survive in an atmosphere that lacked 02?
Ans: During the process of respiration, oxygen is utilised, and carbon dioxide, water and energy are released as products. The combustion reaction requires oxygen. But some cells live where oxygen may or may not be available. There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen. Even among present-day living organisms, we know of several that are adapted to anaerobic conditions.

Q16. It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Ans: Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red fibres. These muscles also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production. These muscles, therefore, can also be called aerobic muscles.

Q17. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.
Ans: The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration because fermentation (anaerobic respiration) accounts
for only a partial breakdown erf glucose, whereas in aerobic respiration it is completely degraded to CO2 and H2O. So, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q18. RuBP carboxylase, PEPcase, Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase. Select/choose enzymes from the list above which are involved in
a. Photosynthesis
b. Respiration
c. Both in photosynthesis and respiration
Ans: a. Photosynthesis: RuBP carboxylase, PEPcase, ATPase
b. Respiration: Hexokinase, ATPase, Pyruvate dehydrogenase, Cytochrome oxidase
c. Both in photosynthesis and respiration: ATPase

Q19. How does a tree trunk exchange gases with the environment although it lacks stomata?
Ans: In stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Q20. Write two energy yielding reactions of glycolysis.
Ans: The conversion of BPGA to 3-phosphoglyceric acid (PGA), is an energy yielding process; this energy is trapped by the formation of ATP. Another ATP is synthesised during the conversion of PEP to pyruvic acid.

Q21. Name the site(s) of pyruvate synthesis. Also, write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.

Q22.Mention the important series of events of aerobic respiration that occurs in the matrix of the mitochondrion as well as one that takes place in the inner membrane of the mitochondrion.
Ans: For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria. The crucial events in aerobic respiration are:
The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of C0₂.

The passing on of the electrons removed as part of the hydrogen atoms to molecular 0₂ with simultaneous synthesis of ATP. What is interesting to note is that the first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.

Q23.Respiratory pathway is believed to be a catabolic pathway. However, nature of TCA cycle is amphibolic. Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Long Answer Type Questions
Q1. In the following flow chart, replace the symbols a,b,c and d with appropriate terms. Briefly explain the process and give any two application of it.

Q2. Given below is a diagram showing ATP synthesis during aerobic respiration, replace the symbols A, B,C,D and E by appropriate terms given in the box.

Q3. Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Q4. Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context.
Ans: The Respiratory Balance Sheet:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages. But this kind of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Q5. Give an account of Glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Ans: Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants, this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage carbohydrates. Sucrose is converted into glucose and fructose by the enzyme, invertase, and these two monosaccharides readily enter the glycolytic pathway. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are same. In glycolysis, a chain of ten reactions,
under the control of different enzymes, takes place to produce pyruvate from glucose. Pyruvic acid is then the key product of glycolysis. The metabolic fate of pyruvate depends on the cellular need. There are three major ways in which different cells handle pyruvic acid produced by glycolysis. These are lactic acid fermentation, alcoholic fermentation and aerobic respiration. Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to C02 and H20, however, organisms adopt Krebs’ cycle which is also called as aerobic respiration. This requires 0₂ supply.

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